Question
The points A(1, 2, 1) , B(−3, 1, 4) , C(5, −1, 2) and D(5, 3, 7) are the vertices of a tetrahedron.
a.Find the vectors \(\overrightarrow {{\text{AB}}} \) and \(\overrightarrow {{\text{AC}}} \).[2]
b.Find the Cartesian equation of the plane \(\prod \) that contains the face ABC.[4]
▶️Answer/Explanation
Markscheme
\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}}
{ – 4} \\
{ – 1} \\
3
\end{array}} \right)\), \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}}
4 \\
{ – 3} \\
1
\end{array}} \right)\) A1A1
Note: Accept row vectors.
[2 marks]
\(\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} = \left| {\begin{array}{*{20}{c}}
{\boldsymbol{i}}&{\boldsymbol{j}}&{\boldsymbol{k}} \\
{ – 4}&{ – 1}&3 \\
4&{ – 3}&1
\end{array}} \right| = \left( {\begin{array}{*{20}{c}}
8 \\
{16} \\
{16}
\end{array}} \right)\) M1A1
normal \({\boldsymbol{n}} = \left( {\begin{array}{*{20}{c}}
1 \\
2 \\
2
\end{array}} \right)\) so \({\boldsymbol{r}} \cdot \left( {\begin{array}{*{20}{c}}
1 \\
2 \\
2
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
1 \\
2 \\
1
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
1 \\
2 \\
2
\end{array}} \right)\) (M1)
\(x + 2y + 2z = 7\) A1
Note: If attempt to solve by a system of equations:
Award A1 for 3 correct equations, A1 for eliminating a variable and A2 for the correct answer.
[4 marks]
Examiners report
Most candidates attempted this question and scored at least a few marks in (a) and (b). Part (c) was more challenging to many candidates who were unsure how to find the required distance. Part (d) was attempted by many candidates some of whom benefited from follow through marks due to errors in previous parts. However, many candidates failed to give the correct answer to this question due to the use of the simplified vector found in (b) showing little understanding of the role of the magnitude of this vector. Part (e) was poorly answered. Overall, this question was not answered to the expected level, showing that many candidates have difficulties with vectors and are unable to answer even standard questions on this topic.
Most candidates attempted this question and scored at least a few marks in (a) and (b). Part (c) was more challenging to many candidates who were unsure how to find the required distance. Part (d) was attempted by many candidates some of whom benefited from follow through marks due to errors in previous parts. However, many candidates failed to give the correct answer to this question due to the use of the simplified vector found in (b) showing little understanding of the role of the magnitude of this vector. Part (e) was poorly answered. Overall, this question was not answered to the expected level, showing that many candidates have difficulties with vectors and are unable to answer even standard questions on this topic.
Question
Consider the points \({\text{A(1, 0, 0)}}\), \({\text{B(2, 2, 2)}}\) and \({\text{C(0, 2, 1)}}\).
A third plane \({\Pi _3}\) is defined by the Cartesian equation \(16x + \alpha y – 3z = \beta \).
a.Find the vector \(\overrightarrow {{\text{CA}}} \times \overrightarrow {{\text{CB}}} \).[4]
b.Find an exact value for the area of the triangle ABC.[3]
c.Show that the Cartesian equation of the plane \({\Pi _1}\), containing the triangle ABC, is \(2x + 3y – 4z = 2\).[3]
d.A second plane \({\Pi _2}\) is defined by the Cartesian equation \({\Pi _2}:4x – y – z = 4\). \({L_1}\) is the line of intersection of the planes \({\Pi _1}\) and \({\Pi _2}\).
Find a vector equation for \({L_1}\).[5]
e.Find the value of \(\alpha \) if all three planes contain \({L_1}\).[3]
f.Find conditions on \(\alpha \) and \(\beta \) if the plane \({\Pi _3}\) does not intersect with \({L_1}\).[2]
▶️Answer/Explanation
Markscheme
\(\overrightarrow {{\rm{CA}}} = \left( \begin{array}{c}1\\ – 2\\ – 1\end{array} \right)\) (A1)
\(\overrightarrow {{\rm{CB}}} = \left( \begin{array}{c}2\\0\\1\end{array} \right)\) (A1)
Note: If \(\overrightarrow {{\text{AC}}} \) and \(\overrightarrow {{\text{BC}}} \) found correctly award (A1) (A0).
\(\overrightarrow {{\rm{CA}}} \times \overrightarrow {{\rm{CB}}} = \left| {\begin{array}{*{20}{c}}i&j&k\\1&{ – 2}&{ – 1}\\2&0&1\end{array}} \right|\) (M1)
\(\left( \begin{array}{c} – 2\\ – 3\\4\end{array} \right)\) A1
[4 marks]
METHOD 1
\(\frac{1}{2}\left| {\overrightarrow {{\text{CA}}} \times \overrightarrow {{\text{CB}}} } \right|\) (M1)
\( = \frac{1}{2}\sqrt {{{( – 2)}^2} + {{( – 3)}^2} + {4^2}} \) (A1)
\( = \frac{{\sqrt {29} }}{2}\) A1
METHOD 2
attempt to apply \(\frac{1}{2}\left| {{\text{CA}}} \right|\left| {{\text{CB}}} \right|\sin C\) (M1)
\({\text{CA.CB}} = \sqrt 5 .\sqrt 6 \cos C \Rightarrow \cos C = \frac{1}{{\sqrt {30} }} \Rightarrow \sin C = \frac{{\sqrt {29} }}{{\sqrt {30} }}\) (A1)
\({\text{area}} = \frac{{\sqrt {29} }}{2}\) A1
[3 marks]
METHOD 1
r.\(\left( \begin{array}{c} – 2\\ – 3\\4\end{array} \right) = \left( \begin{array}{l}1\\0\\0\end{array} \right) \bullet \left( \begin{array}{c} – 2\\ – 3\\4\end{array} \right)\) M1A1
\( \Rightarrow – 2x – 3y + 4z = – 2\) A1
\( \Rightarrow 2x + 3y – 4z = 2\) AG
METHOD 2
\( – 2x – 3y + 4z = d\)
substituting a point in the plane M1A1
\({\text{d}} = – 2\) A1
\( \Rightarrow – 2x – 3y + 4z = – 2\)
\( \Rightarrow 2x + 3y – 4z = 2\) AG
Note: Accept verification that all 3 vertices of the triangle lie on the given plane.
[3 marks]
METHOD 1
\(\left| {\begin{array}{*{20}{c}}{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}}\\2&3&{ – 4}\\4&{ – 1}&{ – 1}\end{array}} \right| = \left( \begin{array}{c} – 7\\ – 14\\ – 14\end{array} \right)\) M1A1
\({\mathbf{n}} = \left( \begin{array}{l}1\\2\\2\end{array} \right)\)
\(z = 0 \Rightarrow y = 0,{\text{ }}x = 1\) (M1)(A1)
\({L_1}:{\mathbf{r}} = \left( \begin{array}{l}1\\0\\0\end{array} \right) + \lambda \left( \begin{array}{l}1\\2\\2\end{array} \right)\) A1
Note: Do not award the final A1 if \(\mathbf{r} =\) is not seen.
METHOD 2
eliminate 1 of the variables, eg x M1
\( – 7y + 7z = 0\) (A1)
introduce a parameter M1
\( \Rightarrow z = \lambda \),
\(y = \lambda {\text{, }}x = 1 + \frac{\lambda }{2}\) (A1)
\({\mathbf{r}} = \left( \begin{array}{l}1\\0\\0\end{array} \right) + \lambda \left( \begin{array}{l}1\\2\\2\end{array} \right)\) or equivalent A1
Note: Do not award the final A1 if \(\mathbf{r} =\) is not seen.
METHOD 3
\(z = t\) M1
write x and y in terms of \(t \Rightarrow 4x – y = 4 + t,{\text{ }}2x + 3y = 2 + 4t\) or equivalent A1
attempt to eliminate x or y M1
\(x,{\text{ }}y,{\text{ }}z\) expressed in parameters
\( \Rightarrow z = t\),
\(y = t,{\text{ }}x = 1 + \frac{t}{2}\) A1
\({\mathbf{r}} = \left( \begin{array}{l}1\\0\\0\end{array} \right) + t \left( \begin{array}{l}1\\2\\2\end{array} \right)\) or equivalent A1
Note: Do not award the final A1 if \(\mathbf{r} =\) is not seen.
[5 marks]
METHOD 1
direction of the line is perpendicular to the normal of the plane
\(\left( \begin{array}{c}16\\\alpha \\ – 3\end{array} \right) \bullet \left( \begin{array}{l}1\\2\\2\end{array} \right) = 0\) M1A1
\(16 + 2\alpha – 6 = 0 \Rightarrow \alpha = – 5\) A1
METHOD 2
solving line/plane simultaneously
\(16(1 + \lambda ) + 2\alpha \lambda – 6\lambda = \beta \) M1A1
\(16 + (10 + 2\alpha )\lambda = \beta \)
\( \Rightarrow \alpha = – 5\) A1
METHOD 3
\(\left| {\begin{array}{*{20}{c}}2&3&{ – 4}\\4&{ – 1}&{ – 1}\\{16}&\alpha &{ – 3}\end{array}} \right| = 0\) M1
\(2(3 + \alpha ) – 3( – 12 + 16) – 4(4\alpha + 16) = 0\) A1
\( \Rightarrow \alpha = – 5\) A1
METHOD 4
attempt to use row reduction on augmented matrix M1
to obtain \(\left( {\begin{array}{*{20}{c}}2&3&{ – 4}\\0&{ – 1}&1\\0&0&{\alpha + 5}\end{array}\left| \begin{array}{c}2\\0\\\beta – 16\end{array} \right.} \right)\) A1
\( \Rightarrow \alpha = – 5\) A1
[3 marks]
\(\alpha = – 5\) A1
\(\beta \ne 16\) A1
[2 marks]
Question
The points A, B, C and D have position vectors a, b, c and d, relative to the origin O.
It is given that \(\mathop {{\text{AB}}}\limits^ \to = \mathop {{\text{DC}}}\limits^ \to \).
The position vectors \(\mathop {{\text{OA}}}\limits^ \to \), \(\mathop {{\text{OB}}}\limits^ \to \), \(\mathop {{\text{OC}}}\limits^ \to \) and \(\mathop {{\text{OD}}}\limits^ \to \) are given by
a = i + 2j − 3k
b = 3i − j + pk
c = qi + j + 2k
d = −i + rj − 2k
where p , q and r are constants.
The point where the diagonals of ABCD intersect is denoted by M.
The plane \(\Pi \) cuts the x, y and z axes at X , Y and Z respectively.
a.i.Explain why ABCD is a parallelogram.[1]
a.ii.Using vector algebra, show that \(\mathop {{\text{AD}}}\limits^ \to = \mathop {{\text{BC}}}\limits^ \to \).[3]
b.Show that p = 1, q = 1 and r = 4.[5]
c.Find the area of the parallelogram ABCD.[4]
d.Find the vector equation of the straight line passing through M and normal to the plane \(\Pi \) containing ABCD.[4]
e.Find the Cartesian equation of \(\Pi \).[3
f.i.Find the coordinates of X, Y and Z.[2]
f.ii.Find YZ.[2]
▶️Answer/Explanation
Markscheme
a pair of opposite sides have equal length and are parallel R1
hence ABCD is a parallelogram AG
[1 mark]
attempt to rewrite the given information in vector form M1
b − a = c − d A1
rearranging d − a = c − b M1
hence \(\mathop {{\text{AD}}}\limits^ \to = \mathop {{\text{BC}}}\limits^ \to \) AG
Note: Candidates may correctly answer part i) by answering part ii) correctly and then deducing there
are two pairs of parallel sides.
[3 marks]
EITHER
use of \(\mathop {{\text{AB}}}\limits^ \to = \mathop {{\text{DC}}}\limits^ \to \) (M1)
\(\left( \begin{gathered}
2 \hfill \\
– 3 \hfill \\
p + 3 \hfill \\
\end{gathered} \right) = \left( \begin{gathered}
q + 1 \hfill \\
1 – r \hfill \\
4 \hfill \\
\end{gathered} \right)\) A1A1
OR
use of \(\mathop {{\text{AD}}}\limits^ \to = \mathop {{\text{BC}}}\limits^ \to \) (M1)
\(\left( \begin{gathered}
– 2 \hfill \\
r – 2 \hfill \\
1 \hfill \\
\end{gathered} \right) = \left( \begin{gathered}
q – 3 \hfill \\
2 \hfill \\
2 – p \hfill \\
\end{gathered} \right)\) A1A1
THEN
attempt to compare coefficients of i, j, and k in their equation or statement to that effect M1
clear demonstration that the given values satisfy their equation A1
p = 1, q = 1, r = 4 AG
[5 marks]
attempt at computing \(\mathop {{\text{AB}}}\limits^ \to \, \times \mathop {{\text{AD}}}\limits^ \to \) (or equivalent) M1
\(\left( \begin{gathered}
– 11 \hfill \\
– 10 \hfill \\
– 2 \hfill \\
\end{gathered} \right)\) A1
area \( = \left| {\mathop {{\text{AB}}}\limits^ \to \, \times \mathop {{\text{AD}}}\limits^ \to } \right|\left( { = \sqrt {225} } \right)\) (M1)
= 15 A1
[4 marks]
valid attempt to find \(\mathop {{\text{OM}}}\limits^ \to = \left( {\frac{1}{2}\left( {a + c} \right)} \right)\) (M1)
\(\left( \begin{gathered}
1 \hfill \\
\frac{3}{2} \hfill \\
– \frac{1}{2} \hfill \\
\end{gathered} \right)\) A1
the equation is
r = \(\left( \begin{gathered}
1 \hfill \\
\frac{3}{2} \hfill \\
– \frac{1}{2} \hfill \\
\end{gathered} \right) + t\left( \begin{gathered}
11 \hfill \\
10 \hfill \\
2 \hfill \\
\end{gathered} \right)\) or equivalent M1A1
Note: Award maximum M1A0 if ‘r = …’ (or equivalent) is not seen.
[4 marks]
attempt to obtain the equation of the plane in the form ax + by + cz = d M1
11x + 10y + 2z = 25 A1A1
Note: A1 for right hand side, A1 for left hand side.
[3 marks]
putting two coordinates equal to zero (M1)
\({\text{X}}\left( {\frac{{25}}{{11}},\,0,\,0} \right),\,\,{\text{Y}}\left( {0,\,\frac{5}{2},\,0} \right),\,\,{\text{Z}}\left( {0,\,0,\,\frac{{25}}{2}} \right)\) A1
[2 marks]
\({\text{YZ}} = \sqrt {{{\left( {\frac{5}{2}} \right)}^2} + {{\left( {\frac{{25}}{2}} \right)}^2}} \) M1
\( = \sqrt {\frac{{325}}{2}} \left( { = \frac{{5\sqrt {104} }}{4} = \frac{{5\sqrt {26} }}{2}} \right)\) A1
[4 marks]
Examiners report
[N/A]
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Question
The parallelogram \(ABCD\) has vertices \(A(3,2,0), B(7,-1,-1), C(10,-3,0)\) and \(D(6,0,1)\). Calculate the area of the parallelogram.
▶️Answer/Explanation
Ans:
\(A(3,2,0), B(7,-1,-1), C(10,-3,0), D(6,0,1)\)
Two bounding vectors are \(\overrightarrow{AB}= \begin{pmatrix} 7\\ -1\\ -1\end{pmatrix} – \begin{pmatrix} 3\\ 2\\ 0\end{pmatrix} = \begin{pmatrix} 4\\ -3\\ -1\end{pmatrix}\) and
\(\overrightarrow{AD}= \begin{pmatrix} 6\\ 0\\ 1\end{pmatrix} – \begin{pmatrix} 3\\ 2\\ 0\end{pmatrix} = \begin{pmatrix} 3\\ -2 \\ 1\end{pmatrix}\)
Area of Parallelogram = \(\left | \begin{pmatrix} i & j & k\\ 4 & -3 & -1\\ 3 & -2 & 1 \end{pmatrix} \right | \)
\(=|(-3-2)i-(4+3)j-(-8+9)k|=\sqrt{(-5)^2+(-7)^2+(-1)^2}\)
\(=\sqrt{75} (accept 5\sqrt{3} or 8.66)\)
Question
Consider the points \(A(1, 2, –4), B(1, 5, 0)\) and \(C(6, 5, –12)\). Find the area of △\(ABC\).
▶️Answer/Explanation
Ans:
METHOD 1
\(\overrightarrow{BA}= \begin{pmatrix} 0\\ -3\\ -4\end{pmatrix}, \overrightarrow{BC}= \begin{pmatrix} 5\\ 0\\ -12\end{pmatrix}\)
Note: Award \((A1), (A1)\) for any two correct vectors used to find area.
\(\overrightarrow{BA} × \overrightarrow{BC} = \begin{pmatrix} i & j & k\\ 0 & -3 & -4\\ 5 & 0 & -12 \end{pmatrix}\)
\(= 36i – 20j + 15k\)
Area = \(\frac{1}{2}|\overrightarrow{BA} × \overrightarrow{BC}|=\frac{1}{2}\sqrt{(36)^2+(20)^2+(15)^2}\)
\(=\frac{1}{2}\sqrt{1921}\)
\(= 21.9 \)
METHOD 2
\(\overrightarrow{BA}= \begin{pmatrix} 0\\ -3\\ -4\end{pmatrix}, \overrightarrow{BC}= \begin{pmatrix} 5\\ 0\\ -12\end{pmatrix}\)
\(|\overrightarrow{BA}|= 5\) \(|\overrightarrow{BC}|= 13\)
Area = \(\frac{1}{2} × 5 × 13 sin \left (cos^{-1}\left (\frac{48}{65}\right ) \right ) = 21.9\)
Question
Given that \(a=2i-j-k, b=2i+j-2k\) and \(c=-i+j-k\) are the position vectors of the points A, B and C respectively, calculate the area of triangle \(ABC\).
▶️Answer/Explanation
Ans:
METHOD 1
\(\overrightarrow{AB}=2j-k\) and \(\overrightarrow{AC}=-3i+2j\)
\(\overrightarrow{AB} × \overrightarrow{AC}= \begin{pmatrix} i & j & k\\ 0 & 2 & -1\\ -3 & 2 & 0 \end{pmatrix}\)
\(=2i+3j+6k\)
Area △ \(ABC\) \(= \frac{1}{2}|\overrightarrow{AB} × \overrightarrow{AC}|\)
\(=\frac{7}{2}\)
METHOD 2
\(\overrightarrow{AB}=2j-k\) and \(\overrightarrow{AC}=-3i+2j\)
Attempting to use the scalar product to find \(\theta \) i.e \(\overrightarrow{AB}.\overrightarrow{AC}=|\overrightarrow{AB}||\overrightarrow{AC}| cos \theta\)
\(cos \theta = \frac{4}{\sqrt{65}}\left ( \theta = arc cos\frac{4}{\sqrt{65}}=arcsin \frac{7}{\sqrt{65}}=60.255…\right )\)
Area △ \(ABC\) \(=\frac{1}{2}|\overrightarrow{AB}| |\overrightarrow{AC}| sin \theta \left ( =\frac{1}{2} ×\sqrt{5}×\sqrt{13}×\frac{7}{\sqrt{65}} \right )\)
\(=\frac{7}{2}\)
Question
Given any two non-zero vectors \(a\) and \(b\), show that \(│a × b│ ^2 = │a│ ^2│b│ ^2 – (a • b) ^2\).
▶️Answer/Explanation
Ans:
METHOD 1
Use of \(| a × b | = | a | | b | sin \theta \)
\(| a × b | ^2 = | a |^2 | b |^2 sin^2 \theta\)
Note: Only one of the first two marks can be implied.
\(= | a |^2 | b |^2 (1 − cos^2 \theta)\)
\(= | a |^2 | b |^2 − | a |^2 | b |^2 cos^2 \theta\)
\(= | a |^2 | b |^2 − (| a | | b | cosθ)^ 2\)
Note: Only one of the above two A1 marks can be implied.
\(= | a |^2 | b |^2 − (a • b)^ 2 \)
Hence LHS = RHS
METHOD 2
Use of \(a • b = | a | | b | cos \theta \)
\(| a |^2 | b |^2 − (a • b)^ 2 = | a |^2 | b |^2 − (| a | | b | cos\theta)^ 2\)
\(= | a |^2 | b |^2 − | a |^2 | b |^2 cos^2 \theta\)
Note: Only one of the above two A1 marks can be implied.
\(= | a |^2 | b |^2 (1 − cos^2 \theta) \)
\(= | a |^2 | b |^2 sin^2 \theta= | a × b |^2\)
Hence LHS = RHS